If-the-roots-of-ax-2-bx-c-0-are-one-another-s-cube-then-show-that-b-2-2ac-2-ac-a-c-2-

Question Number 205527 by MATHEMATICSAM last updated on 23/Mar/24

If the roots of {a x}^{2} + b x + c = 0 are one

{another}^{'} s cube then show that

{(b^{2} - 2 a c)}^{2} = a c {(a + c)}^{2} .

Answered by A5T last updated on 23/Mar/24

L e t t h e r o o t s b e α a n d β . T h e n, α = β^{3} a n d β = α^{3}

α + β = α + α^{3} = \frac{- b}{a}; α β = α^{4} = \frac{c}{a} \Rightarrow α = \underline{+} \sqrt[4]{\frac{c}{a}}

\Rightarrow \underline{+} \sqrt[4]{\frac{c}{a}} \underline{+} \sqrt[4]{\frac{c^{3}}{a^{3}}} = \frac{- b}{a} \Rightarrow \sqrt{\frac{c}{a}} + \sqrt{\frac{c^{3}}{a^{3}}} + \frac{2 c}{a} = \frac{b^{2}}{a^{2}}

\Rightarrow \frac{c}{a} + \frac{c^{3}}{a^{3}} + \frac{2 c^{2}}{a^{2}} = \frac{{(b^{2} - 2 a c)}^{2}}{a^{4}}

\Rightarrow a^{3} c + {a c}^{3} + 2 a^{2} c^{2} = a c {(a + c)}^{2} = {(b^{2} - 2 a c)}^{2}

Answered by Rasheed.Sindhi last updated on 24/Mar/24

AnOther Way

{(b^{2} - 2 a c)}^{2} = a c {(a + c)}^{2}

\Leftrightarrow {(a^{2} (\frac{b^{2}}{a^{2}} - \frac{2 a c}{a^{2}}))}^{2} = a c {(a (\frac{a}{a} + \frac{c}{a}))}^{2}

\Leftrightarrow a^{4} {({(- \frac{b}{a})}^{2} - 2 (\frac{c}{a}))}^{2} = a^{3} c {(1 + \frac{c}{a})}^{2}

\Leftrightarrow {({(- \frac{b}{a})}^{2} - 2 (\frac{c}{a}))}^{2} = \frac{a^{3} c}{a^{4}} {(1 + \frac{c}{a})}^{2}

\Leftrightarrow {({(- \frac{b}{a})}^{2} - 2 (\frac{c}{a}))}^{2} = \frac{c}{a} {(1 + \frac{c}{a})}^{2}

\Leftrightarrow {({(α + β)}^{2} - 2 (α β))}^{2} = α β {(1 + α β)}^{2}

\Leftrightarrow {(α^{2} + β^{2})}^{2} = α β (1 + 2 α β + α^{2} β^{2})

\Leftrightarrow α^{4} + 2 α^{2} β^{2} + β^{4} = α β + 2 α^{2} β^{2} + α^{3} β^{3}

α^{3} = β, β^{3} = α \Rightarrow α^{3} β^{3} = α β

α^{4} = α \cdot α^{3} = α β, β^{4} = β^{3} \cdot β = α β

\Leftrightarrow α β + 2 α^{2} β^{2} + α β = α β + 2 α^{2} β^{2} + α β

\Leftrightarrow 2 α β + 2 α^{2} β^{2} = 2 α β + 2 α^{2} β^{2}

H e n c e p r o v e d

Answered by Rasheed.Sindhi last updated on 24/Mar/24

Still another way

α + β = α + α^{3} = - \frac{b}{a} \Rightarrow \begin{matrix} b = - a α^{3} - a α \end{matrix}

α β = α α^{3} = α^{4} = \frac{c}{a} \Rightarrow \begin{matrix} c = a α^{4} \end{matrix}

{(b^{2} - 2 a c)}^{2} = a c {(a + c)}^{2}

\Rightarrow {({(- a α^{3} - a α)}^{2} - 2 a (a α^{4}))}^{2} = a (a α^{4}) {(a + a α^{4})}^{2}

{\Rightarrow (a^{2} {(α^{3} + α)}^{2} - 2 a^{2} α^{4}))}^{2} = a^{2} (a^{2} α^{4}) {(1 + α^{4})}^{2}

{\Rightarrow a^{4} ((α^{6} + 2 α^{4} + α^{2}) - 2 α^{4}))}^{2} = (a^{4} α^{4}) {(1 + α^{4})}^{2}

{\Rightarrow a^{4} ((α^{6} + α^{2})))}^{2} = a^{4} α^{4} {(1 + α^{4})}^{2}

\Rightarrow a^{4} (α^{12} + 2 α^{8} + α^{4} = a^{4} α^{4} (1 + 2 α^{4} + α^{8})

\Rightarrow a^{4} (α^{12} + 2 α^{8} + α^{4}) = a^{4} (α^{4} + 2 α^{8} + α^{12})

H e n c e p r o v e d